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DENIUS [597]
3 years ago
11

A certain factory whistle can be heard up to a distance of 2.5 km. Assuming that the acoustic output of the whistle is uniform i

n all directions, how much acoustic power is emitted by the whistle? The threshold of human hearing is 1.0 × 10-12 W/m2.
Physics
1 answer:
enyata [817]3 years ago
4 0

Answer:

Emitted power will be equal to 7.85\times 10^{-5}watt

Explanation:

It is given factory whistle can be heard up to a distance of R=2.5 km = 2500 m

Threshold of human hearing I=10^{-12}W/m^2

We have to find the emitted power

Emitted power is equal to P=I\times A

P=I\times 4\pi R^2

P=10^{-12}\times 4\times 3.14\times  2500^2=7.85\times 10^{-5}watt

So emitted power will be equal to 7.85\times 10^{-5}watt

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What is meant by fundamental unit? write any two difference between mass and weight.​
yulyashka [42]

Answer:

Fundamental unit is any unit that is not dependent on other units and other units can be derived from them

Explanation:

Units such as Kilogram, Mass and Time are said to be fundamental units because they are independent.

Differences between Mass and weight;

1. Mass is the measure of the amount of matter in a body while weight is a measure of how the force of gravity acts upon that mass.

2. Mass is a scalar quantity while weight is a vector quantity

8 0
2 years ago
The 20-g bullet is travelling at 400 m/s when it becomes embedded in the 2-kg stationary block. The coefficient of kinetic frict
nikklg [1K]

Answer:

The distance the block will slide before it stops is 3.3343 m

Explanation:

Given;

mass of bullet, m₁ = 20-g = 0.02 kg

speed of the bullet, u₁ =  400 m/s

mass of block, m₂ = 2-kg

coefficient of kinetic friction,  μk = 0.24

Step 1:

Determine the speed of the bullet-block system:

From the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the bullet-block system after collision

(0.02 x 400) + (2 x 0) = v (0.02 + 2)

8 = v (2.02)

v = 8/2.02

v = 3.9604 m/s

Step 2:

Determine the time required for the bullet-block system to stop

Apply the principle of conservation momentum of the system

v(m_1+m_2) -F_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -N \mu_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -g(m_1 +m_2) \mu_kt = v_f(m_1 +m_2)\\\\3.9604(2.02)-9.8(2.02)0.24t = v_f(2.02)\\\\8 - 4.751t = 2.02v_f\\\\3.9604 - 2.352t = v_f

when the system stops, vf = 0

3.9604 -2.352t = 0

2.352t = 3.9604

t = 3.9604/2.352

t = 1.684 s

Thus, time required for the system to stop is 1.684 s

Finally, determine the distance the block will slide before it stops

From kinematic, distance is the product of speed and time

S = \int\limits {v} \, dt \\\\S = \int\limits^t_0 {(3.9604-2.352t)} \, dt\\\\ S = 3.9604t - 1.176t^2

Now, recall that t = 1.684 s

S = 3.9604(1.684) - 1.176(1.684)²

S = 6.6693 - 3.3350

S = 3.3343 m

Thus, the distance the block will slide before it stops is 3.3343 m

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Explanation:

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